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Step-by-Step Solution
Step 1: Rewrite the integral in a form suitable for substitution
We start with the given integral:
\displaystyle \int \frac{dx}{(x^2 + x + 1)^2} .
Observe that x^2 + x + 1 can be written as
\left(x + \tfrac12\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 .
Hence,
(x^2 + x + 1)^2 = \Bigl[\left(x + \tfrac12\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2\Bigr]^2.
Step 2: Make the trigonometric substitution
Set
x + \tfrac12 = \tfrac{\sqrt{3}}{2} \tan \theta .
Then,
dx = \tfrac{\sqrt{3}}{2}\sec^2 \theta \, d\theta .
Step 3: Rewrite the integral in terms of \theta
Substitute into the integral:
\[
\int \frac{dx}{\bigl(x^2 + x + 1\bigr)^2}
= \int \frac{\tfrac{\sqrt{3}}{2} \sec^2 \theta \, d\theta}{\Bigl[\tfrac{\sqrt{3}}{2}\tan \theta\Bigr]^2 + \Bigl(\tfrac{\sqrt{3}}{2}\Bigr)^2} \times \text{(correction in denominator for the square)}.
\]
More carefully, in the denominator we have
\[
\Bigl[\bigl(x + \tfrac12\bigr)^2 + \bigl(\tfrac{\sqrt{3}}{2}\bigr)^2\Bigr]^2
= \Bigl[\bigl(\tfrac{\sqrt{3}}{2}\tan \theta\bigl)^2 + \bigl(\tfrac{\sqrt{3}}{2}\bigr)^2\Bigr]^2
= \Bigl[\tfrac{3}{4}\tan^2 \theta + \tfrac{3}{4}\Bigr]^2
= \Bigl(\tfrac{3}{4}\bigl(\tan^2\theta + 1\bigr)\Bigr)^2
= \tfrac{9}{16}\sec^4 \theta.
\]
Thus,
\[
\int \frac{dx}{(x^2 + x + 1)^2}
= \int \frac{\tfrac{\sqrt{3}}{2} \sec^2 \theta \, d\theta}{\tfrac{9}{16} \sec^4 \theta}
= \frac{\tfrac{\sqrt{3}}{2}}{\tfrac{9}{16}} \int \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta
= \frac{8}{3\sqrt{3}} \int \cos^2 \theta \, d\theta.
\]
Step 4: Integrate using the identity for \cos^2 \theta
Recall that
\cos^2 \theta = \frac{1 + \cos(2\theta)}{2} .
So,
\[
\int \cos^2 \theta \, d\theta
= \int \frac{1 + \cos(2\theta)}{2} \, d\theta
= \frac{1}{2} \int (1 + \cos(2\theta)) \, d\theta
= \frac{1}{2} \Bigl(\theta + \tfrac12 \sin(2\theta)\Bigr) + C.
\]
Hence,
\[
\frac{8}{3\sqrt{3}} \int \cos^2 \theta \, d\theta
= \frac{8}{3\sqrt{3}} \times \frac{1}{2} \Bigl(\theta + \tfrac12 \sin(2\theta)\Bigr) + C
= \frac{4}{3\sqrt{3}} \Bigl(\theta + \tfrac12 \sin(2\theta)\Bigr) + C.
\]
Step 5: Convert back to x
We used
x + \tfrac12 = \tfrac{\sqrt{3}}{2} \tan \theta \implies \tan \theta = \frac{2x+1}{\sqrt{3}}.
Thus,
\theta = \tan^{-1}\Bigl(\frac{2x+1}{\sqrt{3}}\Bigr).
Also,
\sin(2\theta) = \frac{2\tan \theta}{1+\tan^2 \theta},
so
\sin(2\theta) = \frac{2\bigl(\tfrac{2x+1}{\sqrt{3}}\bigr)}{1 + \bigl(\tfrac{2x+1}{\sqrt{3}}\bigr)^2}
= \frac{\tfrac{2(2x+1)}{\sqrt{3}}}{1 + \tfrac{(2x+1)^2}{3}}
= \frac{\tfrac{2(2x+1)}{\sqrt{3}}}{\tfrac{3 + (2x+1)^2}{3}}
= \frac{6(2x+1)}{\sqrt{3}\bigl[3 + (2x+1)^2\bigr]}.
\]
But note that
x^2 + x + 1 = x^2 + x + \tfrac14 + \tfrac34
= \bigl(x + \tfrac12\bigr)^2 + \tfrac34
= \bigl(\tfrac{\sqrt{3}}{2}\tan \theta\bigr)^2 + \bigl(\tfrac{\sqrt{3}}{2}\bigr)^2
= \tfrac{3}{4}(\tan^2 \theta + 1)
= \tfrac{3}{4}\sec^2 \theta.
Hence,
\[
\frac{2x+1}{x^2 + x + 1}
= \frac{2x+1}{\tfrac{3}{4}\sec^2 \theta}
= \frac{4(2x+1)}{3} \cos^2 \theta.
\]
Combining all terms carefully yields the expression in the form:
\[
\int \frac{dx}{(x^2 + x + 1)^2}
= a \tan^{-1}\Bigl(\frac{2x+1}{\sqrt{3}}\Bigr)
+ b \frac{2x+1}{x^2 + x + 1}
+ C.
\]
Comparing with the derived expression shows
\[
a = \frac{4}{3\sqrt{3}}, \quad b = \frac{1}{3}.
\]
Step 6: Compute 9\bigl(\sqrt{3}a + b\bigr)
\[
9\bigl(\sqrt{3}a + b\bigr)
= 9\Bigl(\sqrt{3} \times \frac{4}{3\sqrt{3}} + \frac{1}{3}\Bigr)
= 9\Bigl(\frac{4}{3} + \frac{1}{3}\Bigr)
= 9 \times \frac{5}{3}
= 15.
\]
Final Answer
The value of 9(\sqrt{3}a + b) is \boxed{15} .
